Mokbel Karam, and Prof. Tony Saad (www.tsaad.net)
Department of Chemical Engineering
University of Utah
from IPython.display import Image
from IPython.core.display import HTML
In this Jupyter notebook we will execute the code presented in the paper.
Using mass, length and time as fundamental physical dimensions:
from buckinghampy import BuckinghamPi
Pressure_In_Bubble = BuckinghamPi()
Pressure_In_Bubble.add_variable(name='{\\Delta}p',units='M*L^(-1)*T^(-2)') # pressure
Pressure_In_Bubble.add_variable(name='R',units='L') # diameter
Pressure_In_Bubble.add_variable(name='\\sigma',units='M*T^(-2)') # surface tension
try:
Pressure_In_Bubble.generate_pi_terms()
Pressure_In_Bubble.print_all()
except Exception as e:
print(e)
The number of variables has to be greater than the number of physical dimensions.
Using force and length as fundamental physical dimensions:
from buckinghampy import BuckinghamPi
Pressure_In_Bubble = BuckinghamPi()
Pressure_In_Bubble.add_variable(name='{\\Delta}p',units='F*L^(-2)') # pressure
Pressure_In_Bubble.add_variable(name='R',units='L') # diameter
Pressure_In_Bubble.add_variable(name='\\sigma',units='F*L^(-1)') # surface tension
Pressure_In_Bubble.generate_pi_terms()
Pressure_In_Bubble.print_all()
sets Pi 1 ------ -------------------------- 1 \frac{\sigma}{R {\Delta}p}
from buckinghampy import BuckinghamPi
Pressure_Drop = BuckinghamPi()
Pressure_Drop.add_variable(name='{\\Delta}p',units='M*L^(-1)*T^(-2)') # pressure drop
Pressure_Drop.add_variable(name='R',units='L') # length of the pipe
Pressure_Drop.add_variable(name='d',units='L') # diameter of the pipe
Pressure_Drop.add_variable(name='\\mu',units='M*L^(-1)*T^(-1)') # viscosity
Pressure_Drop.add_variable(name='Q',units='L^(3)*T^(-1)') # volumetic flow rate
Pressure_Drop.generate_pi_terms()
Pressure_Drop.print_all()
sets Pi 1 Pi 2 ------ ------------------------------------------------------- ------------------------------------------------------- 1 \frac{R}{d} \frac{Q \mu}{d^{3} {\Delta}p} 2 \frac{R \sqrt[3]{{\Delta}p}}{\sqrt[3]{Q} \sqrt[3]{\mu}} \frac{d \sqrt[3]{{\Delta}p}}{\sqrt[3]{Q} \sqrt[3]{\mu}} 3 \frac{d}{R} \frac{Q \mu}{R^{3} {\Delta}p}
from buckinghampy import BuckinghamPi
Economic_Growth = BuckinghamPi()
Economic_Growth.add_variable(name='P',units='K',explicit=True) # capital
Economic_Growth.add_variable(name='L',units='Q/T') # labor per period of time
Economic_Growth.add_variable(name='{\\omega_{L}}',units='K/Q') # wages per labor
Economic_Growth.add_variable(name='Y',units='K/T') # profit per period of time
Economic_Growth.add_variable(name='r',units='1/T') # rental rate period of time
Economic_Growth.add_variable(name='{\\delta}',units='1/T') # depreciation rate
Economic_Growth.generate_pi_terms()
Economic_Growth.print_all()
sets Pi 1 Pi 2 Pi 3 ------ --------------------------------- ------------------------ ------------------ 1 \frac{P r}{L {\omega_{L}}} \frac{Y}{L {\omega_{L}}} \frac{{\delta}}{r} 2 \frac{P {\delta}}{L {\omega_{L}}} \frac{Y}{L {\omega_{L}}} \frac{r}{{\delta}} 3 \frac{P r}{Y} \frac{L {\omega_{L}}}{Y} \frac{{\delta}}{r} 4 \frac{P {\delta}}{Y} \frac{L {\omega_{L}}}{Y} \frac{r}{{\delta}}